We know-
[tex]\text{$D=b^{2}-4ac$}[/tex]
-and we know its property.
[tex]\Large\text{$\boxed{\text{If $D=0$, the two roots are identical.}}$}[/tex]
[tex]\;[/tex]
Let us replace the three variables by [tex]\text{$(2p+1),-(7p+2),(7p-3)$}[/tex].
[tex]\text{$D=(7p+2)^{2}-4(2p+1)(7p-3)$}[/tex]
[tex]\text{$D=(49p^{2}+28p+4)-4(14p^{2}+p-3)$}[/tex]
[tex]\text{$D=(49p^{2}+28p+4)-(56p^{2}+4p-12)$}[/tex]
[tex]\Large\text{$\boxed{D=-(7p^{2}-24p-16)}$}[/tex]
[tex]\text{$7p^{2}-24p-16=0$}[/tex]
[tex]\text{$(7p+4)(p-4)=0$}[/tex]
[tex]\Large\text{$\boxed{\text{$7p+4=0$ or $p-4=0$}}$}[/tex]
There arise two cases.
[tex]\text{[Case 1] $\dots$ $p=-\dfrac{4}{7}$}[/tex]
[tex]\text{[Case 2] $\dots$ $p=4$}[/tex]
Firstly,
[tex]\Large\text{$\boxed{\text{$x=\dfrac{\alpha+\beta}{2}$ as two roots are identical.}}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{7p+2}{2p+1}\cdot\dfrac{1}{2}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{49p+14}{14p+7}\cdot\dfrac{1}{2}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{-28+14}{-8+7}\cdot\dfrac{1}{2}$}[/tex]
[tex]\Large\text{$\boxed{\text{$\dfrac{\alpha+\beta}{2}=7$}}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{28+2}{8+1}\cdot\dfrac{1}{2}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{30}{9}\cdot\dfrac{1}{2}$}[/tex]
[tex]\Large\text{$\boxed{\text{$\dfrac{\alpha+\beta}{2}=\dfrac{5}{3}$}}$}[/tex]
So,
[tex]\Large\text{$\boxed{\text{$x=7$ (double solution)}}$}[/tex]
[tex]\Large\text{$\boxed{\text{$x=\dfrac{5}{3}$ (double solution)}}$}[/tex]
Find the values of p for which the equadratic equation (2p + 1) x2- (7p + 2) x + (7p - 3) = 0 has equal roots. Also find these roots.
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Verified answer
We know-
[tex]\text{$D=b^{2}-4ac$}[/tex]
-and we know its property.
[tex]\Large\text{$\boxed{\text{If $D=0$, the two roots are identical.}}$}[/tex]
[tex]\;[/tex]
Let us replace the three variables by [tex]\text{$(2p+1),-(7p+2),(7p-3)$}[/tex].
[tex]\text{$D=(7p+2)^{2}-4(2p+1)(7p-3)$}[/tex]
[tex]\text{$D=(49p^{2}+28p+4)-4(14p^{2}+p-3)$}[/tex]
[tex]\text{$D=(49p^{2}+28p+4)-(56p^{2}+4p-12)$}[/tex]
[tex]\Large\text{$\boxed{D=-(7p^{2}-24p-16)}$}[/tex]
[tex]\;[/tex]
[tex]\text{$7p^{2}-24p-16=0$}[/tex]
[tex]\text{$(7p+4)(p-4)=0$}[/tex]
[tex]\Large\text{$\boxed{\text{$7p+4=0$ or $p-4=0$}}$}[/tex]
[tex]\;[/tex]
There arise two cases.
[tex]\text{[Case 1] $\dots$ $p=-\dfrac{4}{7}$}[/tex]
[tex]\text{[Case 2] $\dots$ $p=4$}[/tex]
[tex]\;[/tex]
Firstly,
[tex]\Large\text{$\boxed{\text{$x=\dfrac{\alpha+\beta}{2}$ as two roots are identical.}}$}[/tex]
[tex]\;[/tex]
[tex]\text{[Case 1] $\dots$ $p=-\dfrac{4}{7}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{7p+2}{2p+1}\cdot\dfrac{1}{2}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{49p+14}{14p+7}\cdot\dfrac{1}{2}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{-28+14}{-8+7}\cdot\dfrac{1}{2}$}[/tex]
[tex]\Large\text{$\boxed{\text{$\dfrac{\alpha+\beta}{2}=7$}}$}[/tex]
[tex]\;[/tex]
[tex]\text{[Case 2] $\dots$ $p=4$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{7p+2}{2p+1}\cdot\dfrac{1}{2}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{28+2}{8+1}\cdot\dfrac{1}{2}$}[/tex]
[tex]\text{$\dfrac{\alpha+\beta}{2}=\dfrac{30}{9}\cdot\dfrac{1}{2}$}[/tex]
[tex]\Large\text{$\boxed{\text{$\dfrac{\alpha+\beta}{2}=\dfrac{5}{3}$}}$}[/tex]
[tex]\;[/tex]
So,
[tex]\text{[Case 1] $\dots$ $p=-\dfrac{4}{7}$}[/tex]
[tex]\Large\text{$\boxed{\text{$x=7$ (double solution)}}$}[/tex]
[tex]\;[/tex]
[tex]\text{[Case 2] $\dots$ $p=4$}[/tex]
[tex]\Large\text{$\boxed{\text{$x=\dfrac{5}{3}$ (double solution)}}$}[/tex]
Question :-
Find the values of p for which the equadratic equation (2p + 1) x2- (7p + 2) x + (7p - 3) = 0 has equal roots. Also find these roots.
Explanation :-
[tex]\sf\red{▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂}[/tex]